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# Which of the following is correct –IValence electronic configurationIISuccessive Ionization energya$n{s^1}$p19,27,36,48,270b$n{s^2}$q16,28,34,260c$n{s^2}n{p^1}$r18,26,230,250d$n{s^2}n{p^2}$s14,200,220,240A. (a)-r, (b)-p , (c)-q ,(d)-sB. (a)-p, (b)-s , (c)-q ,(d)-rC. (a)-q, (b)-s , (c)-r ,(d)-pD. (a)-s, (b)-r , (c)-q ,(d)-p

Last updated date: 13th Jun 2024
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Hint: A comparison between successive ionization energies answers this question.
Ionization energy is directly proportional to the number of valence electrons.
The more tightly bound electrons are difficult to remove and hence the ionization enthalpy increases.

We first need to understand the electronic configuration
To better understand electronic configuration, we take an example here,
${}_6^{12}C$ i.e. Carbon atom (Atomic Number $=$ 6 and Atomic Mass number $=$12)
Atomic number refers to the number of protons in an atom and atomic mass number refers to the total number of protons and neutrons in an atom.
Hence the 6 electrons (equal to the number of protons) of carbon are distributed in the atom as follows –
${C_6}^{12} = 1{s^2}2{s^2}2{p^2}$
Now we understand the valence shell,
The valence shell is the outermost orbital shell and the electrons present in it are the valence electrons.
In the above example,
The last orbital shell comprises two subshells, $2{s^2}$ and $2{p^2}$, and by adding the number of electrons present in these subshells, we get the total number of electrons present in the last shell i.e. the total number of valence electrons.
Hence, the number of valence electrons in a carbon atom is $2 + 2 = 4$ electrons
Hence, the valence shell electronic configuration would be the representation of these valence electrons and it would be $2{s^2}2{p^2}$
Ionization energy deals with the minimum amount of energy required to remove 1 electron from an atom.
Similarly, First Ionization energy deals with the minimum amount of energy required to remove 1 electron from a neutral atom.
Second Ionization energy deals with the minimum amount of energy required to remove 1 electron from a singly charged atom.
Third Ionization energy deals with the minimum amount of energy required to remove 1 electron from a doubly charged atom.
And so on ….
The second ionization energy is always greater than the first ionization energy is the energy required to remove an electron from a cation is more as the electron is now more tightly bound by the nucleus.
For $n{s^1}$,
There is 1 valence electron,
$\therefore I.{E_2} > > I.{E_1}$
In option (s) 14,200,220,240
The energy gap between $I.{E_1}$ and $I.{E_2}$ is comparatively greater.
Hence, (a) $\to$ s
(B) For n$n{s^2}$,
There are 2 valence electrons,
So, $I.{E_3} > > I.{E_2}$
In option (r) 18, 26,230,250
The energy gap between $I.{E_3}$and $I.{E_2}$ is comparatively greater.
Hence, (b) $\to$ r
(C) For $n{s^2}n{p^1}$,
There are 3 valence electrons,
$\therefore I.{E_4} > > I.{E_3}$
In option (q) 16, 28, 34,260
The energy gap between $I.{E_4}$and $I.{E_3}$ is comparatively greater.
Hence, (c) $\to$ q
(D) For $n{s^2}n{p^2}$,
There are 4 valence electrons,
$\therefore I.{E_5} > > I.{E_4}$
In option (p) 19,27,36,48,270
The energy gap between $I.{E_5}$and $I.{E_4}$is comparatively greater. Hence, (d) $\to$ p
 IValence electronic configuration IISuccessive Ionization energy a $n{s^1}$ s 14,200,220,240 b $n{s^2}$ r 18,26,230,250 c $n{s^2}n{p^1}$ q 16,28,34,260 d $n{s^2}n{p^2}$ p 19,27,36,48,270

So, the correct answer is Option D.

Note: One enthalpy each from one row which makes a set of ionization enthalpies for a particularly given atom. From left to right in a period, the first ionization energy increases. The electronic configuration works on the Aufbau principle. It describes the pattern of distribution of electrons in an atom.